A) Plate
B) Sphere
C) Cube
D) None of these
Correct Answer: A
Solution :
Rate of cooling \[\frac{\Delta \theta }{t}=\frac{A\varepsilon \sigma ({{T}^{4}}-T_{0}^{4})}{mc}\]Þ \[\frac{\Delta \theta }{t}\propto A\]. Since area of plate is largest so it will cool fastest.
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